5377_reduce_binary_number.py

'''
Given a number s in their binary representation. Return the number of steps to reduce it to 1 under the following rules:

If the current number is even, you have to divide it by 2.
If the current number is odd, you have to add 1 to it.
It's guaranteed that you can always reach to one for all testcases.

Example 1:

Input: s = "1101"
Output: 6
Explanation: "1101" corressponds to number 13 in their decimal representation.
Step 1) 13 is odd, add 1 and obtain 14.
Step 2) 14 is even, divide by 2 and obtain 7.
Step 3) 7 is odd, add 1 and obtain 8.
Step 4) 8 is even, divide by 2 and obtain 4.
Step 5) 4 is even, divide by 2 and obtain 2.
Step 6) 2 is even, divide by 2 and obtain 1.
Example 2:

Input: s = "10"
Output: 1
Explanation: "10" corressponds to number 2 in their decimal representation.
Step 1) 2 is even, divide by 2 and obtain 1.
Example 3:

Input: s = "1"
Output: 0
'''
class Solution:
    def numSteps(self, s: str) -> int:
        num = int(s, 2)
        steps = 0
        while num != 1:
            if num % 2 == 0:
                num = num // 2
            else:
                num += 1
            steps += 1
        return steps


s = Solution()
string = "10"
res = s.numSteps(string)
print(res)